Change of variable in integral of e^(-nx²)

[Castilla]

I am trying to follow a proof of
$$\int_{0}^{+\infty}e^{-x^2}dx = \frac{\sqrt{\pi}}{2}$$ but the first impasse I find is that, "with the change of variable" $$\sqrt{n}x = y$$ they justify this equality:
$${\frac{1}{\sqrt{n}}\int_{0}^{+\infty}e^{-y^2}dy = \int_{0}^{+\infty}e^{-nx^2}dx$$.

Maybe you can help me to see how they did it? Thanks.

 

[Galileo]

$$\int_{0}^{+\infty}e^{-x^2}dx$$
is just the half of
$$\int_{-\infty}^{+\infty}e^{-x^2}dx$$

The second is usually done with Fubini's theorem and polar substitution.

 

[Castilla]

Er... but the book from where I take the problem only deals with functions of one real variable... they don't use Fubini...

 

[VietDao29]

$$\sqrt{n}x = y$$, so, you have: $$d(\sqrt{n}x) = \sqrt{n}dx = dy$$
Then, you have:
$$\frac{1}{\sqrt{n}} \int_{0} ^ {+ \infty} e ^ {-y^2} dy = \frac{1}{\sqrt{n}} \int_{0} ^ {+\infty} e ^ {-nx ^ 2} (\sqrt{n} dx)$$
$$= \sqrt{n} \times \frac{1}{\sqrt{n}} \int_{0} ^ {+\infty} e ^ {-nx ^ 2} dx = \int_{0} ^ {+\infty} e ^ {-nx ^ 2} dx$$.
Can you get it now?
Viet Dao,

 

[Castilla]

Yes, Viet Dao. Really thanks.